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For, the two triangles ABD, ADC having the same altitude are to each other as their bases (Geom., Bk. IV., P. 6, C.): hence, the triangle is divided into parts having the ratio of m to n.

II. To run a line parallel to one side of a triangular field, that shall form with the parts of the two other sides a


triangle equivalent to the part of the field.


40. Let CBA represent a triangular field and CA the side parallel to which the dividing line is to be drawn.

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At D, erect the perpendicular DG to the diameter BC, and draw BG. Then, with B as a centre, and BG as a radius, describe the arc of a circle cutting BC at E Through E draw EF parallel to CA, and it will divide the triangle in the required ratio.


For, (Geom., Bk. IV., P. 23.)


BE=BOX BC; whence,


BE2 : BC2 :: BD : BC :: m : n.
But, since the triangles BEF, BCA are similar,

Wherefore, from equality of ratios,

BEF BCA :: m : n;



BEF: =

× BCA.


REMARK. The points E and F may easily be found by computation.


For, since BE2=BCX BD, and BD=x BC,

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Let it be required to divide the triangular field CAB, in which AC-9 ch. AB= 11 ch. and CB=7 ch. into two such parts that ADE shall be one-fourth of the whole field.

In this case, we have

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AE-4 ch. 50 1. and AD=5 ch. 50 1.

III. To run a line from a given point in the boundary of a piece of land, so as to cut off, on either side of the line, a given portion of the field.

41. Make a complete survey of the field, by the rules already given. Let us take, as an example, the field whose area is computed at page 118. That field contains 1044 1R 16P, and the following is a plot of it.

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Let it now be required to run a line from station A, in such a manner as to cut off on the left any part of the

field; say,

26A 2R 31 P.

It is seen, by examining the field, that the division line will probably terminate on the course CD. Therefore, draw a line from A to C, which we will call the first closing line.

The bearings and lengths of the courses AB, BC, are always known; and in the present example are found in the table on page 118: hence, the bearing and distance from C to A, can be calculated by Art. 35: they are in this example,

Bearing S 9° 28' E: Course 22.8 ch.

Having calculated the bearing and length of the closing line, find, by the general method, the area which it cuts off: that area, in the present case, is

13A 3R 3P.

It is now evident that the division line must fall on the right of the closing line AC, and must cut off an area ACH, equal to the difference between that already cut off, and the given area: that is, an area equal


26A 2R 31P given area,

13A 3R 3P area already cut off,

12A 3R 28 P.

Since the bearing of the next course CD, and the bearing of the closing line AC are known, the angle ACD which they form with each other, can be calculated, and is in this example 80° 32°. Hence, knowing the hypothenuse AC, and the angle ACG at the base, the length AG of the perpendicular let fall on the course CD, can be found, and is 22.49 chains.

Since the area of a triangle is equal to its base multiplied by half its altitude, it follows, that the base is equal to the area divided by half the altitude. Therefore, if the


12A 3R 28P

be reduced to square chains, and divided by 11.244 chains, which is half the perpendicular AG, the quotient, which is 11.58 chains, will be the base CH. Hence, if we lay off

then run the line AH, it will cut off from the land the required area.

REMARK I. If the part cut off by the first closing line, should exceed the given area, the division line will fall on the left of AC.

REMARK II. If the difference between the given area and the first area' cut off, divided by half the perpendicular AG, gives a quotient larger than the course CD; then, draw a line from A to D, and consider it as the first closing line, and let fall a perpendicular on DE.

REMARK III. When the point from which the division line is to be drawn, falls between the extremities of a course, dividing the course into two parts, consider one of the parts as an entire course, and the other as forming a new course, having the same bearing. The manner of making the calculation will then be the same as before. '



1. Soon after the organization of the present government, several of the states ceded to the United States large tracts of wild land, and these, together with the lands since acquired by treaty and purchase, constitute what is called. the public lands, or public domain. Previous to the year 1802, these lands were parcelled out without reference to any general plan, in consequence of which the titles often. conflicted with each other, and in many cases, several grants covered the same premises.

In the year 1802, the following method of surveying the public lands, was adopted by Colonel Jared Mansfield, then surveyor-general of the North-Western Territory.

2. The country to be surveyed is first divided by

again, by a system of east and west lines, also six miles from each other. The country is thus divided into equal squares, which are called townships. Hence, each township is a square, six miles on a side, and contains thirty-six

square miles.

3. For the purpose of illustration, we have obtained from the general land office the accompanying map, which represents a considerable portion of the State of Arkansas.

The principal meridian in this Survey is called the 5th meridian, and passes through the point of junction of the White river and the Mississippi. The principal base line, running east and west, intersects this meridian a little to the east of White river; and from the meridian and base line, reckoned from this point of intersection, all the ranges of townships are laid off.

For example, 1 North, will apply to all the townships lying in the first row north of the base line: 1 South, will apply to all the townships in the first row south of the base line. Range 1 East, will apply to all the townships lying in the first row, east of the 5th meridian: and range 1 West, will apply to all lying in the first row to the west of it. The small figures designate the rows of townships, reckoned north and south from the base line, and the ranges reckoned east and west from the 5th meridian. Thus, township 1 North, range 4 West, has its exact place designated, and may be immediately located.

4. The principal meridians, and the principal base lines are established by astronomical observation, and the lines of subdivision run with the compass.

For convenience in making surveys, and for the purpose of designating particular localities, a state or large tract, is often divided into parts called "Districts." There are three such districts in the map before us, the boundaries of which are designated by the full dark lines.

5. Each township is divided into equal squares, by meridians one mile apart, and by east and west lines at the same distance from each other. Hence, each township is

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